In triangle $ABC$ let $A'$, $B'$, $C'$ respectively be the midpoints of the sides $BC$, $CA$, $AB$. Furthermore let $L$, $M$, $N$ be the projections of the orthocenter on the three sides $BC$, $CA$, $AB$, and let $k$ denote the nine-point circle. The lines $AA'$, $BB'$, $CC'$ intersect $k$ in the points $D$, $E$, $F$. The tangent lines on $k$ in $D$, $E$, $F$ intersect the lines $MN$, $LN$ and $LM$ in the points $P$, $Q$, $R$. Prove that $P$, $Q$ and $R$ are collinear.
Problem
Source: MMC 2014 Problem 4
Tags: geometry, trigonometry, projective geometry, trig identities, Law of Sines, geometry proposed
aZpElr68Cb51U51qy9OM
09.06.2014 04:54
By Menelaus' Theorem, it is equivalent to prove that $\frac{LR}{MR} \cdot \frac{NQ}{LQ} \cdot \frac{MP}{NP} = 1.$ By Power of Point and Law of Sines, \[\begin{aligned} \left(\frac{RL}{RM} \cdot \frac{QN}{QL} \cdot \frac{PM}{PN}\right)^2 = 1 & \iff \frac{RL \cdot RM}{RM^2} \cdot \frac{QL \cdot QN}{QL^2} \cdot \frac{PM \cdot PN}{PN^2} = 1 \\&\iff \frac{RF}{RM} \cdot \frac{QE}{QL} \cdot \frac{PD}{PN} = 1 \\& \iff \frac{\sin \angle RMF}{\sin \angle RFM} \cdot \frac{\sin \angle QLE}{\sin \angle QEL} \cdot \frac{\sin \angle PND}{\sin \angle PDN} = 1 \\& \iff \frac{\sin \angle LMF}{\sin \angle MLF} \cdot \frac{\sin \angle NLE}{\sin \angle ENL} \cdot \frac{\sin \angle MND}{\sin \angle DMN} = 1 \\&\iff \frac{FL}{FM} \cdot \frac{EN}{EL} \cdot \frac{DM}{DN} = 1. \end{aligned}\] Since $\triangle CA'C' \sim \triangle CFL$ and $\triangle CB'C' \sim \triangle CFM$, we have $FL = \frac{CF}{CA'} \cdot C'A'$ and $FM = \frac{CF}{CB'} \cdot C'B'$. So we have to prove that $\left(\frac{C'A'}{CA'} \cdot \frac{CB'}{C'B'}\right) \cdot \left(\frac{A'B'}{AB'} \cdot \frac{AC'}{A'C'}\right) \cdot \left(\frac{B'C'}{BC'} \cdot \frac{BA'}{B'A'}\right) = 1$, which is true.
BSJL
13.06.2014 18:31
First, notice that we need to prove that $ \frac{FR}{MR}\cdot\frac{DP}{NP}\cdot\frac{EQ}{LQ}= 1 $ and this is equivalent to the following statement: thee lines $ DL,EM,FN $ meet at a point. Now, Let circle $ l $ be tangent to $ AB,AC $ ,and circle $ k $ then by using D'Alembet Theorem, we have that $ D,X(55),L $ are on a line $ \Rightarrow $ thee lines $ DL,EM,FN $ meet at a point.
mathuz
22.06.2014 13:51
by Mixtilinear circles $ \Rightarrow $ it's equivalent to $LD,ME,NF$ are concurrent.
gobathegreat
22.06.2014 14:44
mathuz wrote: by Mixtilinear circles $ \Rightarrow $ it's equivalent to $LD,ME,NF$ are collinear. You probably think concurrent?
mathuz
22.06.2014 16:15
ow, yes-yes!
IDMasterz
27.06.2014 16:36
BSJL wrote: First, notice that we need to prove that $ \frac{FR}{MR}\cdot\frac{DP}{NP}\cdot\frac{EQ}{LQ}= 1 $ and this is equivalent to the following statement: thee lines $ DL,EM,FN $ meet at a point. Now, Let circle $ l $ be tangent to $ AB,AC $ ,and circle $ k $ then by using D'Alembet Theorem, we have that $ D,X(55),L $ are on a line $ \Rightarrow $ thee lines $ DL,EM,FN $ meet at a point. Can you elaborate on this. For general interest, this problem can be generalised nicely - it is a purely projective relationship and a proof arises from using pascal and pappus (or just pascal twice).
TelvCohl
29.10.2014 18:33
My solution: Easy to see $ A', B', C' $ are the midpoints of arc $ NLM $, arc $ LMN $, arc $ MNL $ , respectively. Since $ A, B, C $ are three excenters of $ \triangle LMN $ so we get $ D, E, F $ are the tangent points of three Mixtilinear excircles of $ \triangle LMN $ with $ (LMN) $ . Since its well known that $ LD, ME, NF $ are concurrent at $ X(55) $ of $ \triangle LMN $ , so from Jerabek theorem we get $ P, Q, R $ are collinear . Q.E.D